10147. 「一本通 5.1 例 1」石子合并

#include<algorithm>
#include<bitset>
#include<complex>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<climits>
#include<clocale>
#include<cmath>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define E 200010
using namespace std;
inline int read(){
    int res=0,f=1;char ch=getchar();
    while(ch<'0'ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') res=res*10+ch-'0',ch=getchar();
    return res*f;
}
inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x<10) putchar(x+'0');
    else{
        write(x/10);
        putchar(x%10+'0');
    }
}
int n,f[210][210],a[210],sum[210],ans;
int main(){
    n=read();
    for(int i=1;i<=n;i++) a[i]=read(),a[i+n]=a[i];
    for(int i=1;i<=n*2;i++) sum[i]=sum[i-1]+a[i];
    memset(f,63,sizeof(f));
    for(int i=1;i<=n*2;i++) f[i][i]=0;
    for(int L=2;L<=n;L++){
        for(int i=1;i<=n*2-L+1;i++){
            int j=i+L-1;
            for(int k=i;k<j;k++){
                f[i][j]=min(f[i][j],f[i][k]+f[k+1][j]+sum[j]-sum[i-1]);
            }
        }
    }
    ans=2e9;
    for(int i=1;i<=n;i++){
        ans=min(ans,f[i][i+n-1]);
    }
    write(ans);putchar('\n');
    memset(f,0,sizeof(f));
    for(int i=1;i<=n*2;i++) f[i][i]=0;
    for(int L=2;L<=n;L++){
        for(int i=1;i<=n*2-L+1;i++){
            int j=i+L-1;
            for(int k=i;k<j;k++){
                f[i][j]=max(f[i][j],f[i][k]+f[k+1][j]+sum[j]-sum[i-1]);
            }
        }
    }
    ans=0;
    for(int i=1;i<=n;i++){
        ans=max(ans,f[i][i+n-1]);
    }
    write(ans);putchar('\n');
    return 0;
}